The required expression is obtained by subtracting 2b²-a² from b²-2a²
b² - 2a² - (2b² - a²) = b² - 2a² + (-2b² + a²)
= b² - 2a² - 2b² + a²
= (1 - 2)b² + (-2 + 1)a² = -b² - a²
So -b² - a² must be added

Given:
A = 2a² - 4b-1; B = 5a² + 3b - 8 and C = 2a² + 9b + 3
A + B + C = (2a² - 4b - 1) - (5a² + 3b-  8)+(2a² - 9b + 3)
= 2a² - 4b - 1 + (5a² - 3b + 8) + 2a - 9b + 3
= 2a²- 4b - 15a ²- 3b + 8 + 2a² - 9b + 3
= 2a² - 5a² + 2a² - 4b - 3b - 9b -1 + 8 + 3
= (2 - 5 + 2)a² - 4b - 3b - 9b - 1 + 8 + 3
= -a² - 16b + 10

Since we do not know the number of students who get down at the first stop, let us take the number as x. The number of students get down at the second bus stop is 2x.
x + 2x + 6 + 3 = 60
(1 + 2) x + 9 = 60
3x + 9 = 60
3x + 9 – 9 = 60 − 9 [Subtract 9 on both sides]
3x = 51
\(\frac { 3x }{ 3 } =\frac { 51 }{ 3 } \) [Divide by 3 on both sides]
Therefore, x = 17.
Thus, the number of students got down in the first bus stop is 17.

The numbers are natural and consecutive. Let the numbers be x and x + 1.
Given that,
x + (x + 1) = 75
2x + 1 = 75
2x + 1 − 1 = 75 − 1 [Subtract 1 on both sides]
2x + 0 = 74
\(\frac {2x }{2 } =\frac { 74 }{ 2 } \) [Divide 2 at both sides]
x = 37 and x + 1 = 38
Therefore, the required numbers are 37 and 38.

(i) \(\cfrac { 1 }{ 2 } \left( a+b \right) \)
(ii) p²+ q²

We have 6x, -5x, x, 16x are like terms
6y, y, 7y, are like terms
6, - 5, 1,3 are like terms

(i) 4k from 12k
12k - 4k = (12-4)k = 8k
(ii) 15q from 25q
25q - 15q (25 - 15) q = 10q
(iii) 7xyz from 17xyz
17xyz -7xyz = (17 -7) xyz = 10xyz

(i) 7x, -8x, -12x are like terms containing same algebraic variable 'x'.
(ii) 5y, 12 -9y are like terms containing same algebraic variable 'y'.
(ili) 6z, z and 11 z are like terms containing same algebraic variables 'z'.

3x - 5 - x + 9 = 3(3) - 5 - 3 + 9
= 9 - 5 - 3 + 9 =18 - 8 =10

(i) 13x + 12y - 5 from 27x + 5y - 43
27x + 5y-43 - (13x + 12y- 5)
= 27x + 5y - 43 - 13x - 12y + 5
= (27 - 13) x + (5 - 12) y + (- 43) + 5
= 14x + (-7)y + (-38)
= 14x - 7y - 38
(ii) 3p + 5 from p - 2q + 7
= P - 2q + 7 + (- 3p -5)
= (1-3)p-  2q + 2
= -2p - 2q  + 2
(iii) m + n from 3m - 7n
Additive inverse of (m + n) is -m -n,
= 3m -7n - (m + n) = 3m -7n + (-m -n)
= (3m - m) + (-7n - n)
= (3 - 1)m + (-7 -1) n
= 2m - 8n
(iv) 2y + Z from 6z - 5y
Additive inverse of 2y + z = -2y - z.
6z - 5y-(2y + z)
= ( 6z -5y) +(-2y - z)
= (62 - z) + (-5y - 2y)
= (6 - 1)z + (-5 -2)y
= 5z + (-7)y = 5z - 7y

i) (3x − 4y + z) + (2x − z + 3y)
= (3x + 2x) + ( −4y + 3y) + (z − z)
= (3 + 2)x + ( −4 + 3)y + (1 − 1)z
= 5x − 1y + 0z
= 5x − y.

ii) (4x + 3y) − (2x − 5y)
= (4x + 3y) + ( −2x + 5y)
= (4x − 2x) + (3y + 5y)
= (4 − 2)x + (3 + 5)y
= 2x + 8y.